$\sum\limits_{n=2}^{\infty} \dfrac{1}{n\sqrt{\ln(n)}}$ When applying the integral test, we get a limit that determines whether the series converges or diverges. What is this limit? Choose 1 answer: Choose 1 answer: (Choice A) A $\lim_{b\to\infty} \sqrt{\ln(b)}$ (Choice B) B $\lim_{b\to\infty} \left[ 2\sqrt{\ln(b)} - 2\sqrt{\ln(2)} \right]$ (Choice C) C $\lim_{b\to\infty} \left[ \ln \left(b\sqrt{\ln(b)} \right) - \ln \left(2\sqrt{\ln(2)} \right) \right]$ (Choice D) D $\lim_{b\to\infty} \dfrac{1}{b\sqrt{\ln(b)}}$
Solution: $\dfrac{1}{n\sqrt{\ln(n)}}$ satisfies the conditions for the integral test. This means that $\sum\limits_{n=2}^{\infty} \dfrac{1}{n\sqrt{\ln(n)}}$ converges/diverges together with $\int_2^{\infty} \dfrac{1}{x\sqrt{\ln(x)}} \,dx$. $\int_2^{\infty} \dfrac{1}{x\sqrt{\ln(x)}} \,dx=\lim_{b\to\infty} \left[ 2\sqrt{\ln(b)} - 2\sqrt{\ln(2)} \right]$ In conclusion, the limit that determines whether the series converges or diverges is $\lim_{b\to\infty} \left[ 2\sqrt{\ln(b)} - 2\sqrt{\ln(2)} \right]$.